What is the integral of (x^2-3x+8)/((x^2-4x+7)^2) ?

The integral of (x^2-3x+8)/((x^2-4x+7)^2) is

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integral of (x^2-3x+8)/((x^2-4x+7)^2)

\int \frac{x ^{2} - 3 x + 8}{( x ^{2} - 4 x + 7 ) ^{2}} d x

\frac{1}{6 3} x arctan (\frac{1}{3} (x - 2)) + \frac{1}{12 3} x sin (2 arctan (\frac{1}{3} (x - 2))) + \frac{7}{6 3} arctan (\frac{1}{3} (x - 2)) + \frac{1}{12 3} sin (2 arctan (\frac{1}{3} (x - 2))) - \frac{( x - 2 ) ( arctan ( \frac{1}{3} ( x - 2 ) ) ( x ^{2} - 4 x + 7 ) + 3 ( x - 2 ) )}{6 3 ( x ^{2} - 4 x + 7 )} - \frac{1}{2 ( x ^{2} - 4 x + 7 )} + C

Solution steps

\int \frac{x ^{2} - 3 x + 8}{( x ^{2} - 4 x + 7 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2} - 3 x + 8}{( x ^{2} - 4 x + 7 ) ^{2}} : \frac{1}{x ^{2} - 4 x + 7} + \frac{x + 1}{( x ^{2} - 4 x + 7 ) ^{2}}

= \int \frac{1}{x ^{2} - 4 x + 7} + \frac{x + 1}{( x ^{2} - 4 x + 7 ) ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{x ^{2} - 4 x + 7} d x + \int \frac{x + 1}{( x ^{2} - 4 x + 7 ) ^{2}} d x

\int \frac{1}{x ^{2} - 4 x + 7} d x = \frac{1}{3} arctan (\frac{x - 2}{3})

\int \frac{x + 1}{( x ^{2} - 4 x + 7 ) ^{2}} d x = \frac{1}{12 3} (2 arctan (\frac{x - 2}{3}) + sin (2 arctan (\frac{x - 2}{3}))) (x + 1) - \frac{1}{6} \frac{( x - 2 ) ( arctan ( \frac{1}{3} ( x - 2 ) ) ( x ^{2} - 4 x + 7 ) + 3 ( x - 2 ) )}{3 ( x ^{2} - 4 x + 7 )} + \frac{3}{x ^{2} - 4 x + 7}

= \frac{1}{3} arctan (\frac{x - 2}{3}) + \frac{1}{12 3} (2 arctan (\frac{x - 2}{3}) + sin (2 arctan (\frac{x - 2}{3}))) (x + 1) - \frac{1}{6} \frac{( x - 2 ) ( arctan ( \frac{1}{3} ( x - 2 ) ) ( x ^{2} - 4 x + 7 ) + 3 ( x - 2 ) )}{3 ( x ^{2} - 4 x + 7 )} + \frac{3}{x ^{2} - 4 x + 7}

Simplify

\frac{1}{3} arctan (\frac{x - 2}{3}) + \frac{1}{12 3} (2 arctan (\frac{x - 2}{3}) + sin (2 arctan (\frac{x - 2}{3}))) (x + 1) - \frac{1}{6} \frac{( x - 2 ) ( arctan ( \frac{1}{3} ( x - 2 ) ) ( x ^{2} - 4 x + 7 ) + 3 ( x - 2 ) )}{3 ( x ^{2} - 4 x + 7 )} + \frac{3}{x ^{2} - 4 x + 7} : \frac{1}{6 3} x arctan (\frac{1}{3} (x - 2)) + \frac{1}{12 3} x sin (2 arctan (\frac{1}{3} (x - 2))) + \frac{7}{6 3} arctan (\frac{1}{3} (x - 2)) + \frac{1}{12 3} sin (2 arctan (\frac{1}{3} (x - 2))) - \frac{( x - 2 ) ( arctan ( \frac{1}{3} ( x - 2 ) ) ( x ^{2} - 4 x + 7 ) + 3 ( x - 2 ) )}{6 3 ( x ^{2} - 4 x + 7 )} - \frac{1}{2 ( x ^{2} - 4 x + 7 )}

= \frac{1}{6 3} x arctan (\frac{1}{3} (x - 2)) + \frac{1}{12 3} x sin (2 arctan (\frac{1}{3} (x - 2))) + \frac{7}{6 3} arctan (\frac{1}{3} (x - 2)) + \frac{1}{12 3} sin (2 arctan (\frac{1}{3} (x - 2))) - \frac{( x - 2 ) ( arctan ( \frac{1}{3} ( x - 2 ) ) ( x ^{2} - 4 x + 7 ) + 3 ( x - 2 ) )}{6 3 ( x ^{2} - 4 x + 7 )} - \frac{1}{2 ( x ^{2} - 4 x + 7 )}

Add a constant to the solution

= \frac{1}{6 3} x arctan (\frac{1}{3} (x - 2)) + \frac{1}{12 3} x sin (2 arctan (\frac{1}{3} (x - 2))) + \frac{7}{6 3} arctan (\frac{1}{3} (x - 2)) + \frac{1}{12 3} sin (2 arctan (\frac{1}{3} (x - 2))) - \frac{( x - 2 ) ( arctan ( \frac{1}{3} ( x - 2 ) ) ( x ^{2} - 4 x + 7 ) + 3 ( x - 2 ) )}{6 3 ( x ^{2} - 4 x + 7 )} - \frac{1}{2 ( x ^{2} - 4 x + 7 )} + C

\frac{d}{d x} (\frac{x ^{2} - 2 x + 1}{2 - x})

\int_{0}^{π} 3 cos^{6} (x) d x

\int \frac{1}{t ^{2} t ^{2} - 16} d t

\frac{d}{d x} (e^{\frac{x}{6}})

What is the integral of (x^2-3x+8)/((x^2-4x+7)^2) ?
The integral of (x^2-3x+8)/((x^2-4x+7)^2) is